@harambe Lyman Series starts from ground state (n=1), Balmer Series starts from 1st excited state (n=2). So after emission in Balmer Series the atom is left in state n=2 and must emit another photon to reach ground state. en.wikipedia.org/wiki/Hydrogen_spectral_series
The atom gets excited somehow, say to state n=10. There is no way of knowing which state it will decay to - this is all determined at random. If it decays to state n=2 then the emission line is in the Balmer Series. It will be followed by another emission to reach n=1 ground state.
So the question is - there is jump of electron from a higher state to lower state hence electron is emmitted and this emmision has to fall in balmer so n=2?
@harambe I don't understand your question. Can you rephrase it?
If the atom is excited to any state above n=2 there is no reason why it has to decay to n=2 state.
eg If excited to n=10 there could be several emissions to n=7, 5, 3 then 2. Or it could miss out n=2 altogether. Or it could decay from n=10 to n=1 without stopping at any other levels.
There is a certain probability of each decay from state m to state n.
It is all decided at random, according to the probabilities of each route of decay.
One method is to use/prove the virial theorem: for a potential of the form $V\left(r\right) \propto {r^p}$ (for the hydrogen atom, $p = -1$),
$$
\left<T\right>_n = \frac{p}{2} \left<V\right>_n
$$
for the $n$-th energy eigenstate.
Use this, together with
$$
E_n = \left<E\right>_n = \left<T\right>...
Somebody help: with these very easy question: Find number of significant digits in 0.0005 cm, 600cm ? I don't know how the specified units will change the number of significant digits
@Abcd it must be very easy for you please say
@sammygerbil can you help?
I am confused as leading 0 is never significant v/s when units are mentioned all are significant
The only exception is when there is a trailing decimal point, eg 600.0 cm.
This has 3 or possibly 4 significant digits.
When we write 600 without any indication of uncertainty this usually means the number has been rounded to the nearest hundred. But of course we cannot be sure it has not been rounded to the nearest ten or unit.
When we write 600.0 this implies it is accurate to the nearest unit or possibly tenth of a unit.
Where you take the zero of the PE is just a choice. The PE doesn't have to be zero at infinity we could choose it to be 27 or 365.25 at infinity without affecting the physics.
Now the question says we want to take the PE to be zero for the n=1 state, and that means we have to add +27.2 eV to the PE. This makes the PE zero for the n=1 state and +27.2 eV at infinity.
All we are doing is adding a constant +27.2 eV to all PEs.
You put the whatever it is you want to measure between the jaws and turn the screw until it stops. The you read the rough size off the shaft of the screw. Let me see if I can find a diagram ...
Turn the thimble anticlockwise and that opens the jaws. Open the jaws wide enough to put the wire (or whatever) between them, then turn the thimble clockwise to close the jaws again until they grip the wire.
So at this point the distance between the jaws is the same as the size of the wire. Yes?
@JohnRennie Yes, I dont get this step: then turn the thimble clockwise to close the jaws again until they grip the wire. I mean how it yields the measurement.
@JohnRennie Suppose initially the jaws were closed. So the process is equivalent to: Slowly and slowly making space for the object by turning the screw gauge circular scale?
That screw gauge is just like the one in my diagram. The rough scale on the shaft is measured in 0.5 mm, and one complete turn of the screw = 0.5mm i.e. the scale that goes around the thimble measures an offset of zero to 0.5 mm. OK so far?
So when the jaws are closed the gauge shows 0.45mm, i.e. zero on the shaft plus an offset of 0.45mm on the thimble. That means the scale is miscalibrated and always shows 0.45mm more than the real size.
Well, my line of thought was that, it's either 45mm more than 0 or 5mm less than 50. If we take the first case, none of the options match. So, go for the second one.
Suppose the excited state is n=3. That means it can have the following transitions: 3 to 2 3 to 1 2 to 1 (i.e. the 3 to 2 transition is followed by 2 to 1) So in total a gas in that state can emit three different photons. OK so far?
It's not a great question because it expects you to solve it by trial and error. But it doesn't take that much trial and error. My table above identifies the transition and the value of Z.
@JohnRennie In X-ray crystallography we use X rays because the wavelength of x rays is about the same order as the inter atomic distances in a crystal.What happens if we use more shorter wavelengths?
In Q29 the H atom absorbs the photon and it ionises the atom. The binding energy is 13.6eV, so the kinetic energy of the resulting electron and proton is equal to the photon energy minus 13.6eV.
To a first approximation the proton will remain stationary, so we only need consider the kinetic energy of the electron.
I think what the question means is that the photon uses part of its energy to excite the H atom and a lower energy photon carries on. So for example the photon that carries on could have an energy of $12.4 - E_{12}$ eV.
Or an energy of $12.4 - E_{13}$ eV.
So the transmitted beam contains these two lower energies in addition to original energy of 12.4 eV
If an incoming photon has an original energy of 12.4 eV and it transfers only a part of its energy E to the H atom, then carries on, the transmitted beam will contain photons with an energy of 12.4 - E. i.e. in the transmitted beam will now be photons with a reduced energy/longer wavelength.
And those reduced energy photons have longer wavelengths than the original 12.4eV photon. So the transmitted beam contains photons with three wavelengths: the original 100nm and the two longer wavelengths.
@harambe it's basically the same problem, except that because the neutron and proton have similar masses you have to consider the motion of both particles. So you should work in the centre of mass frame.
As a general rule I would solve the problem keeping everything as symbols. So for example write the initial energy as $E_0$ and the momentum is then $p_0 = \sqrt{2mE_0}$.
In fact, unless I've made a silly mistake switching to the COM frame answers the question immediately ...
Suppose the velocity of the neutron is $v$. Since the neutron and H atom have the same mass (approximately) in the COM frame the velocities of the neutron and H atom are going to be $v/2$ and $-v/2$. Yes?
And $KE \propto v^2$, so in the COM frame both the neutron and H atom have a kinetic energy of $12.5/4 = 3.125$ eV. So the collision energy in the COM frame is $6.25$ eV.
Suppose the neutron energy was 25eV in the lab frame, so in the COM frame the energy is 25/4 = 6.25eV. Since in thr COM frame both the neutron and H atom have this energy the total collision energy is 2*6.25 = 12.5eV. Yes?
So what can happen is that some energy gets used up exciting the H atom to the n=2 state, and the total kinetic energy that the neutron and H atom fly apart with is the original 12.5eV minus the n=1 to 2 excitation energy.
@JohnRennie My book says:In the fcc unit cell,two tetrahedral voids are formed on each of the four nonparallel body diagonals of the cube......What does this mean?...I cant visualize this!...How are the voids exactly present?
@JohnRennie In each 1/8 of the unit cell there is a tetrahedral void.When they say that there are two tetrahedral voids present along a diagonal which two are they talking about among these 8?
@JohnRennie No,I'm asking :There are 8 tetrahedral voids right?...They say that two tetrahedral voids are present along each diagonal.Out of those 8 tetrahedral voids which two are they talking about?..Which two of those 8 are present along any one given diagonal.....Lets say im choosing a diagonal AB ...which two of those voids are present along the diagnoal?
@JohnRennie Please dont mind all my last messgaes....I got it..I was stupid
The wire is generating a sound wave and that sound wave is resonating in the column. For the column to resonate at it's fundamental frequency the wire has to be vibrating at that same frequency. So once you've worked out the frequency of the wave in the column you know the frequency of vibration of the wire is the same.
And you know the wavelength of the wave on the wire ...
Suppose you're pushing a swing, and you don't push at the natural frequency of the swing. That means sometimes you'll be pushing when the swing is moving away from you, and sometimes when the swing is moving towards you.
On average some of your pushes will speed up the swing and some will slow it down. So overall you won't make much difference to the amplitude of the swing.
